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\(=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x-1}\)

\(=\dfrac{x+1-1}{x-1}=\dfrac{x}{x-1}\)

a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)

\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)

\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)

7 tháng 6 2023

` @ \color{Red}{m}`

` \color{lightblue}{Answer}`  

\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x}{x+1}\)

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\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\\ =\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2x-6}{2x\left(x+3\right)}\\ =\dfrac{3x-\left(2x-6\right)}{2x\left(x+3\right)}\\ =\dfrac{3x-2x+6}{2x\left(x+3\right)}\\ =\dfrac{x+6}{2x\left(x+3\right)}\)

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\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\\ =\dfrac{1}{1-x}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{1-x}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1+x-2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1}{1+x}\)

7 tháng 6 2023

\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\left(dkxd:x\ne\pm1\right)\)

\(=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x}{x+1}\)

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\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\left(dkxd:x\ne\pm3;x\ne0\right)\)

\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\)

\(=\dfrac{3x-2\left(x-3\right)}{2x\left(x+3\right)}\)

\(=\dfrac{3x-2x+6}{2x\left(x+3\right)}\)

\(=\dfrac{x+6}{2x^2+6x}\)

==========================

\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\left(dkxd:x\ne\pm1\right)\)

\(=-\dfrac{1}{x-1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-\left(x+1\right)+2x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-x-1+2x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{x+1}\)

3 tháng 4 2023

bổ sung \(ĐKXĐ:\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\\x\ne2\end{matrix}\right.\)

3 tháng 4 2023

A=\(\left(\dfrac{1}{x+1}-\dfrac{1}{x^2-1}\right).\dfrac{x+1}{x-2}\)

   =\(\left(\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{1}{x^2-1}\right).\dfrac{x+1}{x-2}\)

   =\(\left(\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}-\dfrac{1}{x^2-1}\right).\dfrac{x+1}{x-2}\)

   =\(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x+1}{x-2}\) 

   =\(\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-2\right)}\)

   =\(\dfrac{1}{x-1}\)

18 tháng 11 2021

\(ĐK:x\ne\pm1\)

Với \(\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\Leftrightarrow A=\dfrac{x+2}{x^2-1}+\dfrac{x^2}{x+1}=\dfrac{x+2+x^3-x^2}{\left(x-1\right)\left(x+1\right)}\)

Với \(-1< x< 1\Leftrightarrow A=\dfrac{x+2}{1-x^2}+\dfrac{x^2}{x+1}=\dfrac{x+2+x^3-x^2}{\left(x+1\right)\left(1-x\right)}\)

\(B=2x\cdot\dfrac{2}{x}+x^2+2x+1\left(x\ne0\right)=x^2+2x+5\)